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If X is a r.v. that follows that Gaussian distribution with mean μ and variance σ^2, how do I find the distribution function for exp(X) where exp() is the exponent function.
If X follows a Gaussian distribution with mean μ and variance σ², then the distribution of Y = exp(X) is derived from the cumulative distribution function (CDF) of X. The CDF of Y is given by F_Y(α) = P(exp(X) ≤ α), which simplifies to F_X(log(α)). The probability density function (PDF) of Y is f_Y(α) = f_X(log(α)) / α, applicable for all positive α, while the PDF is zero for negative α. This relationship leads to a closed-form expression for the PDF of Y, indicating that Y follows a log-normal distribution.
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